| The stability of carbanions in the following compounds |
| (i)\[RC\equiv \overline{{\ddot{C}}}\] |
(ii) |
| (iii)\[{{R}_{2}}C=\overline{{\ddot{C}}}H\] |
| (iv)\[{{R}_{3}}C-\overline{{\ddot{C}}}{{H}_{2}}\] |
| is in the order |
A) (ii) > (iii) > (iv) > (i)
B) (iv) > (ii) > (iii) > (i)
C) (i) > (iii) > (ii) > (iv)
D) (i) > (ii) > (iii) > (iv)
Correct Answer: D
Solution :
[d]: The stability of the carbanion decreases as the electronegativity of the carbon carrying -ve charge decreases or the hybridisation of carbon carrying -ve charge changes from sp to \[s{{p}^{2}}\] to \[s{{p}^{3}}\]. Thus, \[RC\equiv {{C}^{-}}\]is the most stable while \[{{R}_{3}}C-CH_{2}^{-}\]is the least stable carbanion. Out of \[{{C}_{6}}H_{5}^{-}\] and \[{{R}_{2}}C=C{{H}^{-}};{{R}_{2}}C=C{{H}^{-}}\] is less stable due to +I-effect of the two R groups. Thus, the overall stability decreases in the order: \[RC\equiv {{C}^{-}}>{{C}_{6}}H_{5}^{-}>{{R}_{2}}C=CH_{{}}^{-}>{{R}_{3}}C-CH_{2}^{-}\]
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